Question

Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be two functions defined by $f ( x )=\log _{ e }\left( x ^{2}+1\right)- e ^{- x }+1$ and $g(x)=\frac{1-2 e^{2 x}}{e^{x}}$. Then, for which of the following range of $\alpha$, the inequality $f \left( g \left(\frac{(\alpha-1)^{2}}{3}\right)\right)> f \left( g \left(\alpha-\frac{5}{3}\right)\right)$ holds?

• A. $(2,3)$
• B. $(-2,-1)$
• C. $(1,2)$
• D. $(-1,1)$

Answer 1

Answer: (A)

$ f ( x )=\log _{ e }\left( x ^{2}+1\right)- e ^{- x }+1$
$\Rightarrow f ^{\prime}( x )=\frac{2 x }{ x ^{2}+1}+ e ^{- x }>0 \forall x \in R$
$\Rightarrow f$ is strictly increasing
$g(x)=\frac{1-2 e^{2 x}}{e^{x}}=e^{-x}-2 e^{x} $
$\Rightarrow g^{\prime}(x)=-\left(2 e^{x}+e^{-x}\right) < 0 \forall x \in R$
$\Rightarrow g$ is decreasing
Now $f\left(g\left(\frac{(\alpha-1)^{2}}{3}\right)\right)>f\left(g\left(\alpha-\frac{5}{3}\right)\right)$
$\Rightarrow g \left(\frac{(\alpha-1)^{2}}{3}\right)>g\left(\alpha-\frac{5}{3}\right)$
$\Rightarrow \frac{(\alpha-1)^{2}}{3}<\alpha-\frac{5}{3}$
$\Rightarrow \alpha^{2}-5 \alpha+6 < 0$
$\Rightarrow(\alpha-2)(\alpha-3) < 0$
$\Rightarrow \alpha \in(2,3)$